4x3 = 47 - 15

4x3 = 32

x3 = 32 : 4

x3 = 8

x3 = 23

x = 2.

Vậy x = 2.

 -29.(85-47)-85.(47-29)

=-29.38-85.18

=-1102-1503

=-2605

-15.(23-29)+23-(15-29)

=-15.(-6)+23-(-14)

=90+23+14

=113+14

=127

 -13.(55-29)+23.(15-29)

=-13.26+23.(-14)

=-338+(-322)

=-660

-10.(2+4)-5.(4-3)

=-10.6-5.1

=-60-5

=-65

a) 47.(-147) - 47.(-47)

= 47.(-147 + 47)

= 47 . (-100)

= -4700

b) 24.(15 - 4) + 4.(24 - 15)

= 24.15 - 25.4 + 4.25 - 4.15

= (24 - 4).15 - (25.4 - 4.25)

= 20 . 15

= 300

47(-147) - 47(-47)

= 47(-147 + 47)

= 47.(-100)

= -4700

a) 47.(-147)-47.(-47)

=47.(-147+47)

=47x(-100)

=-4700

b) 24.(15-4)+4.(24-15)

=24.11+4.9

=264+4.9

=264+36

=300

\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)

\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=6.\\x=5.\end{matrix}\right.\)

\(b.2x-3x+10=x.\Leftrightarrow-2x+10=0.\Leftrightarrow x=5.\)

\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)

\(d.x^4+x^3-x^2-x=0.\Leftrightarrow x^2\left(x^2+x\right)-\left(x^2+x\right)=0.\Leftrightarrow\left(x^2-1\right)\left(x^2+x\right)=0.\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)x\left(x+1\right)=0.\Leftrightarrow\left(x-1\right)\left(x+1\right)^2x=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x+1=0.\\x=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-1.\\x=0.\end{matrix}\right.\)

\(4.x^3+15=47\)

\(4.x^3=32\)

\(x^3=8\)

\(x^3=2^3\)

\(x=2\)

\(x^3+15=47\)

\(x^3=47-15\)

\(x^3=32\)

\(\Rightarrow\) \(x=?????\)

\(4^x:2+15=47\)

\(\Rightarrow4^x:2=32\)

\(\Rightarrow4^x=64\)

\(\Rightarrow4^x=4^3\)

Vậy \(x=3\)

4^x : 2 + 15 = 47

4^x : 2         = 47 - 15

4^x : 2         = 32

4^x              = 32 x 2

4^x              = 64

4^x              = 4³

 Vậy x       = 3

\(\dfrac{3}{4}+\dfrac{1}{6}\times3+\dfrac{2}{5}:\dfrac{4}{15}\)

\(=\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{2}{5}\times\dfrac{15}{4}\)

\(=\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{3}{2}\) \(=\dfrac{3+2+3}{4}=\dfrac{8}{4}=2\)

Gọi số đó là \(\overline{abcd}\) \(\left(a\ne0\right)\)

Ta có: \(\overline{abcd}+a+b+c+d=2020\)

\(\Leftrightarrow1000a+100b+10c+d+a+b+c+d=2020\)

\(\Leftrightarrow1001a+101b+11c+2d=2020\)